Two resistors, one of 20Ω and the other of 30Ω, are connected in parallel. This combination is connected in series with an 8-Ω resistor and a 12-V battery. The current through the 20-Ω resistor is :

Correct option is C

​$$R_1 = 20\Omega, \quad R_2 = 30\Omega\\\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} \Omega\\R = 12\Omega\\R_3 = 8\Omega\\R = 8 + 12 = 20\Omega\\V = iR\\i = \frac{12}{20} \Rightarrow 0.6 \text{ A}\\\text{Voltage drop across 12 Ohm is } = 12 \times 0.6 = 7.2 \text{ V}\\\text{Current in 20}\Omega \text{ resistor} = \frac{7.2}{20} = 0.36\\= 0.36 \text{ A}$$​​